2009/015) Find a Polynomial with integral coefficients of lowest degree with 2^1/3 + 3^1/2 as a zero

Let x = 2^1/3 + 3^1/2. You want to do enough algebra to make the square and cubes roots go away, until you only have integer (or rational) coefficients.

x - 3^1/2 = 2^1/3

Now, cube both sides:

2 = (x-3^(1/2))^3

we can use the formula

(A - B)^3 = A^3 - 3*A^2*B + 3*A*B^2 - B^3

to get

2 = x^3-3x^2(3^1/2)+9x-3^(3/2) = (x^3+9x -3^3/2(x^2+1)

or 3^(3/2)(x^2+1) = (x^3+9x-2)
now squaring both sides you get

27(x^2+1)^2 = (x^3+9x-2)^2
expanding it we should get the result

27(x^4+2x^2+1) = (x^6 + 81 x^2 + 4 + 18x^4 -4x^3 - 36x)
or 27x^4+54x^2+27 = (x^6 + 81 x^2 + 4 + 18x^4 -4x^3 - 36x)
or (x^6 + 81 x^2 + 4 + 18x^4 -4x^3 - 36x) - (27x^4+54x^2+27) = 0
or x^6 -9 x^4 - 4x^3 + 27x^2 - 36x - 23 = 0 after simplifying